Question

Find the positive value of m for which the distance between the points A (5 -3) B(13 -m) is 10 units

Answer 1

$\pink{d = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } } \\ \\ a = (5 \: \: \: - 3) = (x1 \: \: \: y1) \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ b = (13 \: \: \: - m) = (x2 \: \: \: y2) \: \: \: \: \: \: \: \: \\ \\ 10 = \sqrt{ {(13 - 5)}^{2} + {( - m - ( - 3))}^{2} } \\ \\ 10 = \sqrt{ {8}^{2} + {(3 - m)}^{2} } \\ \\ 100 = {8}^{2} + {(3 - m)}^{2} \\ \\ 100 = 64 + 9 - 6m + {m}^{2} \\ \\ {m}^{2} - 6m + 73 - 100 = 0 \\ \\ {m}^{2} - 6m - 27 = 0 \\ \\ {m }^{2} - 9m + 3m - 27 = 0 \\ \\ m(m - 9) + 3(m - 9) = 0 \\ \\ (m - 9)(m + 3) = 0 \\ \\ m = 9 \: \: \: \: \: or \: \: \: \: \: m = - 3$